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Proof step 26 from proof range-write-in (previous step)

key0 ∈ ri, key ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ ri[key0] = ri[key], ∃ key1. key1 ∈ ri[key, adr] ∧ ri[key, adr][key1] = ri[key0]

User applied rule exists right

And obtained the following new goals

Proof step 27:
key ∈ ri, key0 ∈ ri, ∀ key0, key1. key0 ∈ ri ∧ key1 ∈ ri ∧ ri[key0] = ri[key1] → key0 = key1 ⊦ ri[key0] = ri[key], key0 ∈ ri[key, adr] ∧ ri[key, adr][key0] = ri[key0]

Rule argument

pos: right 2 terms: key0 abort: F discard: T computed: F

Additional information

Used terms: key0